The maximum value of |arg(11−z)| for |z|=1, z≠1 is given by

∵|z|=1. Let z=eiθ⇒11-z=12ieiθ/2⋅sin⁡(θ/2)=−ie−iθ/22sin⁡(θ/2)∴Maximum value of arg⁡11−z=π2

The maximum value of |arg(11−z)| for |z|=1, z≠1 is given by

∵|z|=1. Let z=eiθ⇒11-z=12ieiθ/2⋅sin⁡(θ/2)=−ie−iθ/22sin⁡(θ/2)∴Maximum value of arg⁡11−z=π2

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The maximum value of $| \arg (\frac{1}{1 - z}) | for | z | = 1, z \neq 1$ is given by

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$\frac{π}{2}$

$\frac{π}{6}$

$\frac{π}{3}$

$\frac{2 π}{3}$

answer is C.

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Detailed Solution

$\begin{array}{l} ∵ | z | = 1 . Let z = e^{iθ} \\ \Rightarrow \frac{1}{1 - z} = \frac{1}{2 {ie}^{iθ / 2} \cdot \sin (θ / 2)} = - \frac{{ie}^{- iθ / 2}}{2 \sin (θ / 2)} \end{array}$
$∴$ Maximum value of $|\arg (\frac{1}{1 - z})| = \frac{π}{2}$